Question: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 + x}{x - 2} = \dfrac{4x - 2}{x - 2}$
Multiply both sides by $x - 2$ $ \dfrac{x^2 + x}{x - 2} (x - 2) = \dfrac{4x - 2}{x - 2} (x - 2)$ $ x^2 + x = 4x - 2$ Subtract $4x - 2$ from both sides: $ x^2 + x - (4x - 2) = 4x - 2 - (4x - 2)$ $ x^2 + x - 4x + 2 = 0$ $ x^2 - 3x + 2 = 0$ Factor the expression: $ (x - 2)(x - 1) = 0$ Therefore $x = 2$ or $x = 1$ However, the original expression is undefined when $x = 2$. Therefore, the only solution is $x = 1$.